Anonymous asked in Science & MathematicsChemistry · 1 decade ago

how many grams of Ca(NO3)2 can be produced by reacting excess HNO3 with 7.4g of Ca(OH)2?

3 Answers

  • Anonymous
    1 decade ago
    Favorite Answer

    Ca(OH)2+2HNO3---> Ca(NO3)2 +2H2O

    1 mole + 2moles --- 1mole + 2 moles

    you need to calculate the molecular weight of

    1-Ca(OH)2= 40+2(16+1)=74gr.

    2-Ca(NO3)2=40+2{14+(3x16) }= 164gr

    from the the equation of the reaction :

    Each 74 grams of Ca(OH)2 will produce 164 grams of Ca(NO3)2,

    Based on that 7.4 grams of Ca(OH)2,

    will give=164x7.4/74=


  • maussy
    Lv 7
    1 decade ago

    Ca(OH)2+2HNO3---> Ca(NO3)2 +2H2O

    you see that 1 mole of Ca(OH)2 yields 1 mole Ca(NO3)2

    the molar weight of Ca(OH)2 = 40+2(16+1)=74g

    the molar wight of Ca (NO3)2 =40+2*(14+3*16) =164g

    You see that 7.4g of Ca (OH)2 = 7.4/74=0.1 mole

    you obtain 0.1 mole of Ca (NO3)2 = 0.1*164

    ANSWER =16.4 g

  • 4 years ago

    what is the answer if it reacts with excess HNO3 with 6.48 of Ca(OH)2?

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